1  Geometry of the Complex plane

This section is a brief reminder of Sections 3 and 4 of MA1006 Algebra.

Definition 1.1 The complex numbers \(\C\) are the set of all pairs \(z=(x,y)\in\R^2\) of real numbers with the addition

\[ z_1+z_2=(x_1+x_2,y_1+y_2) \tag{1.1}\]

and the multiplication

\[ z_1\cdot z_2=(x_1x_2-y_1y_2,x_1y_2+x_2y_1), \tag{1.2}\]

where \(z_1=(x_1,y_1)\) and \(z_2=(x_2,y_2).\) We call \(x=\Re(z)\) the real part and \(y=\Im(z)\) the imaginary part of \(z.\)

From Section 3.8 in MA1006 Algebra we recall:

Proposition 1.1 The complex numbers are a field.

Remark 1.1.

We view the real numbers as a subset of \(\C\) by identifying \(x\in\R\) with \((x,0)\in\C.\) The imaginary unit is \(i=(0,1).\) With these conventions, a calculation using Equation 1.2 shows that

\[ z=x+iy. \tag{1.3}\]

Using this notation, we can manipulate complex numbers in the same way as real numbers, keeping in mind the identity

\[ i^2=-1. \tag{1.4}\]

Of course, \(\C\) is a one-dimensional vector space over itself. Restricting the scalar multiplication to \(\R\) makes \(\C\) into a vector space over \(\R,\) isomorphic to \(\R^2,\) of dimension two with standard basis \(1,i\in\C.\)

Proposition 1.2  

1. In the standard basis, every \(A=\begin{pmatrix}a&b \\ c&d\end{pmatrix} \in\M_{2\t2}(\R)\) corresponds uniquely to an \(\R\)-linear map \(T_A\colon\C\to\C,\) namely \[ T_A(x+iy)=(ax+by)+i(cx+dy). \tag{1.5}\]

2. \(T_A\) is \(\C\)-linear \(\iff\) \(a=d\) and \(b=-c.\) In this case, \[ T_A(z)=\al\cdot z,\qquad \al=a+ic. \tag{1.6}\]

Proof.

1. This is a recap from linear algebra. An \(\R\)-linear map \(T\colon\C\to\C\) is uniquely determined by the image of the basis vectors \(T(1),\) \(T(i)\) which, conversely, may be prescribed arbitrarily. If we write \(T(1)=a+ic,\) \(T(i)=b+id,\) then the linear transform \(T\) is described entirely by \(a,b,c,d\in\R.\) The expression Equation 1.5 is obtained by expanding the left hand side by \(\R\)-linearity.

2. In the same way, \(\C\)-linear maps correspond to \(\al\in\M_{1\t1}(\C)\) as in Equation 1.6. An \(\R\)-linear map \(T_A\) is \(\C\)-linear \(\iff\) \(T_A(i)=T_A(i1)=iT_A(1)\) \(\iff\)

   \[\begin{pmatrix}b \\ d \end{pmatrix} =T_A(i)=iT_A(1)=i\begin{pmatrix}a \\ c \end{pmatrix} =\begin{pmatrix}-c \\ a \end{pmatrix}.\]

   In this case, \(T_A(x+iy)=(ax-cy)+i(cx+ay)=\al(x+iy).\)

Definition 1.2 The conjugate of \(z\in\C\) is the complex number \[\ol{z}=(x,-y),\]and the modulus (also called absolute value) is\[|z|=\sqrt{x^2+y^2}\geqslant0.\]

Proposition 1.3 The following formulas hold for \(z,w\in \C:\)

\[\begin{align} \ol{z\cdot w}&=\ol{z}\cdot\ol{w} &\ol{z+w}&=\ol{z}+\ol{w}\\ \ol{\ol{z}}&=z &\ol{i}=-i&,\ \ol{1}=1\\ z\cdot\ol{z}&=|z|^2 &|z\cdot w|&=|z|\cdot|w|\\ \Re(z)&=\frac{z+\ol{z}}{2} &\Im(z)&=\frac{z-\ol{z}}{2i}\\ z^{-1}&=\frac{\ol{z}}{|z|^2}\quad\text{if $z\neq0$} \end{align}\]

Proposition 1.4 The following inequalities hold for \(z,w\in\C:\)

\[|z+w|\leqslant|z|+|w| \qquad |z-w|\geqslant\bigl||z|-|w|\bigr| \tag{1.7}\]

\[|\Re(z)|\leqslant|z| \qquad |\Im(z)|\leqslant|z| \tag{1.8}\]

Thinking of complex numbers as points in the plane, we can use polar coordinates to represent them (see Figure 1.1).

Figure 1.1: Polar coordinates

Proposition 1.5 For every non-zero complex number \(z=(x,y)\) there is an argument \(\th\in\R\) and a radius \(r=|z|>0\) such

\[x=r\cos(\th), y=r\sin(\th). \tag{1.9}\]

This representation is unique up to replacing \(\th\) by \(\th+2\pi k\) for any \(k\in\Z.\)

Proof. (omitted)

Since \(x^2+y^2=r^2(\cos(\th)^2+\sin(\th)^2)=r^2,\) the radius must be \(r=|z|.\) Define the complex number \(w=r^{-1}z\) and write \(w=u+iv\) for its real and imaginary parts.

We will prove the existence of \(\th\in\R\) with \(u=\cos(\th),\) \(v=\sin(\th).\) This also proves the existence of a representation Equation 1.9, by multiplying by \(r.\) Since \(u^2+v^2=|w|^2=r^{-2}|z|^2=1,\) we know \(|u|\leqslant1,\) \(|v|\leqslant1.\) Recall that \(\cos\colon[0,\pi]\to[-1,1]\) and \(\sin\colon[-\pi/2,\pi/2]\to[-1,1]\) are bijections. Hence \[\begin{align*} u&=\cos(\al) &&\text{ for some }\al\in[0,\pi], \\ v&=\sin(\be) &&\text{ for some }\be\in[-\pi/2,\pi/2]. \end{align*}\] As \(\sin(\be)^2=v^2=1-u^2=1-\cos(\al)^2=\sin(\al)^2,\) we have \(\sin(\al)=\pm\sin(\be)=\sin(\pm\be).\) To produce the correct \(\th,\) we distinguish two cases.

Case 1

\(\al\in[0,\pi/2].\) Then \(\al=\pm\be\) by the injectivity of the sine function on the interval \([-\pi/2,\pi/2].\) Setting \(\th=\pm\al=\be,\) we find that \(u=\cos(\th)\) and \(v=\sin(\th),\) as required.

Case 2

\(\al\in[\pi/2,\pi].\) Then \(\pi-\al, \be\in[-\pi/2,\pi/2]\) and \(\sin(\pi-\al)=\sin(\al)=\pm\sin(\be),\) so \(\pi-\al=\pm\be\) by injectivity. Setting \(\th=\pm\al=\pm\pi-\be,\) we find \(u=\cos(\th), v=\sin(\th),\) using trigonometric identities.

This completes the existence part of the proof. For uniqueness, we already know that \(r=|z|>0\) is unique, so it remains to consider \[\begin{align*} x&=r\cos(\th_1)=r\cos(\th_2), &y&=r\sin(\th_1)=r\sin(\th_2). \end{align*}\] To translate the situation into an interval that we understand, pick \(k_1, k_2\in \Z\) so that \(\th_1 + 2\pi k_1, \th_2 + 2\pi k_2 \in [-\pi,\pi).\) Then \[\cos(\th_1+2\pi k_1)=\cos(\th_1)=\cos(\th_2)=\cos(\th_2+2\pi k_2).\]

Using the injectivity of the cosine function and considering cases as above, we find that \(\th_1+2\pi k_1 = \pm(\th_2+2\pi k_2).\) If the sign is ‘\(+\)’ we get \(\th_1-\th_2=2\pi(k_2-k_1)\) and we are done, so suppose \(\th_1+2\pi k_1 = -(\th_2+2\pi k_2).\) Then \[\sin(\th_1)=\sin(\th_2)=\sin(\th_2+2\pi k_1)=-\sin(\th_1+2\pi k_1)=-\sin(\th_1)\]

implies \(\sin(\th_1)=0.\) Therefore \(\th_1\) is a multiple of \(2\pi,\) which implies that \(\th_1+2\pi k_1 = -(\th_2+2\pi k_2)=0\) since these numbers were chosen in \([-\pi,\pi)\) and we have \(2\pi\Z\cap[-\pi,\pi)=\{0\}.\) Hence \(\th_1-\th_2=2\pi(k_2-k_1).\)

To get around the non-uniqueness of the argument in polar coordinates, we restrict \(\th\) to lie in a half-open interval of length \(2\pi.\) Here is the most common convention.

Definition 1.3 The principal argument of a non-zero \(z\in\C\) is the unique \(\th\in(-\pi,\pi]\) such that Equation 1.9 holds, and we write \(\arg(z)=\th.\)

Definition 1.4 The value of the exponential function at the complex number \(z=x+i\th,\) where \(x,\th\in\R,\) is defined as \[e^{x+i\th}=e^x\bigl(\cos(\th)+i\sin(\th)\bigr). \tag{1.10}\]

Proposition 1.5 implies that every complex number can be represented in polar form\[z=re^{i\th}. \tag{1.11}\]

The addition of complex numbers is the usual addition of vectors in \(\R^2.\) To visualize multiplication, the polar form is useful. Combining Equation 1.2 and Equation 1.10, we find that \[\begin{align*} e^{i\th_1}\cdot e^{i\th_2}=&\cos(\th_1)\cos(\th_2)-\sin(\th_1)\sin(\th_2)\\ &+i\bigl[\cos(\th_1)\sin(\th_2)+\cos(\th_2)\sin(\th_1)\bigr] \\ =&\cos(\th_1+\th_2)+i\sin(\th_1+\th_2)=e^{i(\th_1+\th_2)}. \end{align*}\] Here we have used the trigonometric addition formulas. Hence

\[z_1 = r_1e^{i\th_1}, z_2 = r_2e^{i\th_2} \implies z_1z_2 = r_1r_2e^{i(\th_1+\th_2)}. \tag{1.12}\]

Complex multiplication adds the angles and multiplies the radii.

Proposition 1.6 \(e^{z_1}\cdot e^{z_2}=e^{z_1+z_2}\) for all \(z_1, z_2\in\C.\) Moreover, we have \((e^z)^n=e^{nz}\) for all \(z\in\C,\) \(n\in\Z.\)

Proof.

The first assertion follows from Equation 1.12 combined with the identity \(e^{x_1}e^{x_2}=e^{x_1+x_2}\) for \(x_1, x_2\in \R\) from MA1005 Calculus. The second claim follows from this by induction.

The polar form can be applied to the construction of \(n\)^th roots.

For example, the n^th root of unity is \(\ze_n=e^{i\frac{2\pi}{n}}\) and satisfies \[(\ze_n)^n=(e^{i\frac{2\pi}{n}})^n=e^{2\pi i}=1.\]

Proposition 1.7 Every complex number \(z\neq0\) has an \(n\)^th root \(w\) satisfying \(w^n=z.\) If \(w\) is an \(n\)^th root of \(z,\) the set of all \(n\)^th roots of \(z\) is \[\bigl\{w, \ze_n\cdot w, \ze_n^2\cdot w,\ldots, \ze_n^{n-1}\cdot w\bigr\}. \]

Proof.

Write \(z=re^{i\th}\) and \(w=se^{i\varphi}\) for \(\th,\varphi\in[0,2\pi)\) and \(r,s>0.\) By the uniqueness of the polar form, the equation \(w^n=z\) is equivalent to \(s^n=r\) and \(n\varphi=\th+2\pi k\) for some \(k\in\Z.\) Of course, \(s=\sqrt[n]{r}\) is the unique positive \(n\)^th root from MA1005 Calculus. From \(0\leqslant n\varphi<2\pi n\) and \(0<-\th\leqslant2\pi\) we get \[0<k=\frac{n\varphi-\th}{2\pi}<n.\]

Since \(k\) is an integer, this implies \(k=0,\ldots,n-1\) and hence \(\varphi=\frac{\th+2\pi k}{n}\) for such \(k\) are the only possible solutions for \(\varphi.\) In summary, \[w_k = \sqrt[n]{r}e^{i\varphi}=\sqrt[n]{r}e^{i\frac{\th+2\pi k}{n}}=\ze_n^k\cdot w_0, k=0,\ldots,n-1,\]

are all the possible \(n\)^th roots of \(z.\)

Questions for further discussion

• The complex numbers are obtained by ‘adjoining’ a symbol \(i\) with \(i^2=-1.\) If instead we would have adjoined a different symbol \(\ep\) with \(\ep^2=-1,\) would the set of elements \(x+\ep y\) still define a field?
• The real numbers have a total order ‘\(\leqslant\)’. Why doesn’t it make sense to extend this definition to the complex numbers?
• Describe geometrically the set \(R_n=\{1,\ze_n,\ldots,(\ze_n)^{n-1}\}\) of \(n\)^th roots of unity. Find a connection between \(R_n\) and the cyclic group \(C_n=\{\ol{0},\ldots,\ol{n-1}\}\) of integers modulo \(n\) from MX3020 Group Theory.

1.1 Exercises

Note

This problem sheet is intended as a recap and contains more problems than can be discussed during the tutorials.

Exercise 1.1

Verify \[z = x+iy\] and \[i^2 =-1\] straight from the definition Equation 1.2.

Exercise 1.2

How many real solutions \(x\) does \(x^2+1=0\) have? Show that the polynomial equation \(z^2+1=0\) has exactly two solutions \(z\in\C.\)

Exercise 1.3

Give examples of complex numbers \(z,w\neq0\) such that \(z^2+w^2=0.\)

Exercise 1.4

Sketch the position of the complex numbers \(i, 1+i, \frac{3+2i}{4}\) in the plane.

Exercise 1.5

Express the following complex numbers \(z\) in the form \(x+iy\) with \(x,y\in\R.\) \[(1+i)^{20},\ (5+3i)(1+2i),\ (1-i)(2+3i),\ (1-i)i(1+i),\ \frac{2+i}{1-i}\]

Exercise 1.6

Express the following complex numbers \(z\) in the form \(x+iy\) with \(x,y\in\R.\) \[1/i,\ \frac{1}{1+i},\ \frac{3+i}{3-i} \]

Exercise 1.7

Find the modulus and the conjugate of the following complex numbers. \[2+i,\ i,\ 5-3i,\ \frac{1+i}{2+i} \]

Exercise 1.8

Describe the sets \(A=\{z\in\C\mid\Im(z)>0\},\) \(B=\{z\in\C\mid\Re(z)\leqslant1\},\) \(C=\{z\in\C\mid \Re((1+i)z)=0\},\) and \(A\cap B\) geometrically.

Exercise 1.9

Describe the set \(D=\{z\in\C \mid z\cdot\overline{z}=1\}\) geometrically.

Hint: Write \(z=re^{i\th}\) in polar form.

Exercise 1.10

Draw all nine sets described by the following conditions on the complex number \(z.\) \[\begin{align*} |z|&=1,&|z|&<1,&1<&|z|<2,\\ |1+z|&>1,&|2-z|&<2,&3<&|z+i|<4,\\ |z-1|&<|z+1|,&|z|&=|z+1|,&|z-1|&=|z+i|. \end{align*}\]

Exercise 1.11

Let \(S=\{x+iy\in \C \mid 0\leqslant x,y \leqslant 1\}\). Draw \(S\) and the sets \[\begin{align*} A&=\{2z \mid z\in S\}, &B&=\{\ol{z} \mid z\in S\},\\ C&=\{-z \mid z\in S\}, &D&=\{z^2 \mid z\in S\}. \end{align*}\]

Exercise 1.12

Let \(D=\{z\in\C \mid |z|<1\}\) be the unit disk. Draw the sets \[A=\{2z \mid z\in D\},\quad B=\{z^2 \mid z\in D\},\quad C=\{|z| \mid z\in D\}.\]

Exercise 1.13

Show that \(i=e^{i\pi/2}\) and \(-1=e^{i\pi}.\)

Exercise 1.14

Express the following complex numbers \(z\) in the form \(x+iy\) with \(x,y\in\R.\) \[e^{i\pi/4},\ e^{i\pi},\ e^{i\frac{2\pi}{3}} \]

Exercise 1.15

Write each of the following complex numbers in polar form \(re^{i\theta}\) with \(r>0\) and \(-\pi<\theta\leqslant\pi.\) \[i, \quad -1, \quad -i, \quad 1+i, \quad 1-i, \quad i-1, \quad \frac{1}{2}+i\frac{\sqrt{3}}{2}\] Draw each of these numbers in the complex plane.

Exercise 1.16

Calculate \(i^{2021}\) and \((1+i)^{20}.\)

Exercise 1.17

Solve the equation \((1 - i)^n - 2075 = 2021\) and find \(n\in\N.\)

Exercise 1.18

Prove that for \(z\in\R\) we have

\[\begin{align*} \cos(z)&=\frac{e^{iz}+e^{-iz}}{2}, &\sin(z)&=\frac{e^{iz}-e^{-iz}}{2i}. \end{align*}\]

Use these equations to extend the definition of the functions \(\cos(z), \sin(z)\) to complex arguments \(z\in\C.\) Find \(z\in\C\) with \(\sin(z)=2.\)

Hint: Put \(w=e^{iz}\) and reduce to a quadratic equation.

Exercise 1.19

Prove the following statements.

1. \(\ol{zw}=\ol{z}\cdot\ol{w}\) for all \(z,w\in\C\)
2. \(\ol{z_1z_2\cdots z_n} = \ol{z}_1\ol{z}_2\cdots\ol{z}_n\) for all \(z_1, z_2, \ldots, z_n\in\C\) (use induction)
3. \(\ol{(z^n)}=(\ol{z})^n\) for all \(z\in\C\)
4. Let \(p(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_1z+a_0\) be a polynomial with real coefficients \(a_0,\ldots,a_n\in\R.\) Prove that \(\ol{p(z)} = p({\ol{z}}).\) Deduce that all roots of \(p(z)\) occur in complex conjugate pairs.

Exercise 1.20

Show that \(|z|=|{-z}|\) and \(|\ol z|=|z|.\) Prove also that \(|\la z|=\la |z|\) for all \(\la\geqslant 0.\)

Exercise 1.21

Prove that \(\Re(z)=\frac{1}{2}(z+\bar{z}),\) \(\Im(z)=\frac{1}{2i}(z-\bar{z}).\)

Exercise 1.22

Prove that \(\ol{e^z}=e^{\ol z}.\) Deduce that \(|e^z|=e^{\Re(z)}.\)

Exercise 1.23

Prove that \(|z+w|^2+|z-w|^2=2|z|^2+2|w|^2.\)

Exercise 1.24

Show that \(|z+w|^2=|z|^2+2\Re(z\ol w)+|w|^2.\) Use this to determine the conditions on \(z,w\) for \(|z+w|=|z|+|w|\) to hold.

Exercise 1.25

Assuming we know the triangle inequality \(|z+w|\leqslant |z|+|w|\) for all \(z,w\in\C,\) prove the reverse triangle inequality \[|z-w|\geqslant\bigl||z|-|w|\bigr|.\]

Exercise 1.26

Let \(K\) be a field with \(\R\subset K \subset\C.\) Prove that \(K=\R\) or \(K=\C.\)

Further resources

• Freitag–Busam (Freitag and Busam 2009, chap. 1) for additional exercises and historical background.
• https://en.wikipedia.org/wiki/Complex_number for overview and history
• https://youtu.be/T647CGsuOVU for some visualization

Dowson, H. R. 1979. “Serge Lang, Complex Analysis (Addison-Wesley, 1977), Xi + 321 Pp., £1200.” Proceedings of the Edinburgh Mathematical Society 22 (1): 65–65. https://doi.org/10.1017/S0013091500027838.

Freitag, E., and R. Busam. 2009. Complex Analysis. Universitext. Springer Berlin Heidelberg. https://books.google.co.uk/books?id=3xBpS-ZKlgsC.

Howie, J. M. 2012. Complex Analysis. Springer Undergraduate Mathematics Series. Springer London. https://books.google.co.uk/books?id=0FZDBAAAQBAJ.

Jameson, G. J. O. 1987. “H. A. Priestley, Introduction to Complex Analysis (Oxford University Press, 1985), 197 Pp., £8.50.” Proceedings of the Edinburgh Mathematical Society 30 (2): 325–26. https://doi.org/10.1017/S0013091500028406.

Remmert, R. 1991. Theory of Complex Functions. Graduate Texts in Mathematics. Springer New York. https://books.google.co.uk/books?id=uP8SF4jf7GEC.

Thomson, B. S., J. B. Bruckner, and A. M. Bruckner. 2001. Elementary Real Analysis. Prentice-Hall. https://books.google.co.uk/books?id=6l_E9OTFaK0C.